# calculus inequality proof using mean value theorem

Use the mean value theorem to prove that if $$\displaystyle ~|x|<\frac{\pi}2 ~,~~ |y|<\frac{\pi}2~,$$ then $$|\sin(y) - \sin(x)| \leq |y-x| \leq |\tan(y) - \tan(x)|$$

What I did was using the mean value theorem using $$\sin(x)$$ as the function to get $$\frac{ \sin(x) - \sin(y) }{ x-y } = \sin'(c) = \cos(c)$$ for some $$c$$ inbetween $$x$$ and $$y$$, and since the value of $$\cos(x)$$ for every $$x$$ goes around $$1$$ and $$-1$$ $$\left| \frac{ \sin(x) - \sin(y) }{ x-y }\right| \leq 1$$ but I'm stuck here

• So, you have proved the first inequality. Can you use the same idea to prove the right inequality, using MVT for $\tan$? – GReyes May 1 at 2:31

## 1 Answer

Note that if $$f(x)=\tan{x}$$, then by mean value theorem $$\frac{\tan{x}-\tan{y}}{x-y}=\tan'(c)=\sec^2({c})\geq1$$

• ok so with this I get (using the transitive property of inequalities) |sin(y)−sin(x)|≤|y−x|≤|tan(y)−tan(x)| but where does the hypothesis come in play? how do the x|<π/2,|y|<π/2 relate to his? – rorod8 May 1 at 2:40
• in the case of $\tan{x}$ it is necessary to guarantee continuity, for apply the theore – AsdrubalBeltran May 1 at 2:46
• ohhhhhhhh, ok, thank you dude – rorod8 May 1 at 2:47